# Irrational distances, rational area triangles

Given any positive integer $n\ge3$ how to show that there are $n$ distinct points in the plane such that

1- the distance between any two points is irrational number and

2- each set of three points determines a non-degenerate triangle whose area is a rational number

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And what has that $\,n\geq 3\,$ to do with this question at all? – DonAntonio Jul 11 '12 at 21:32
3 because the question mentions triangles and we need at least three points to form triangles. – Frank Jul 11 '12 at 21:33
It's not really a combinatorial statement. – tomasz Jul 11 '12 at 21:36
Geometrical then? – Frank Jul 11 '12 at 21:39
Are you looking for $n$ points? – Sigur Jul 11 '12 at 21:40

Look at the points $(x,x^2)$ where $x$ is a non-negative integer. These are lattice points, so the area of any triangle they determine is rational, indeed of the form $k/2$, where $k$ is an integer.
Now calculate the distance between $(a,a^2)$ and $(b,b^2)$, where $b\gt a$. The square of the distance is $$(b-a)^2+(b^2-a^2)^2,$$ which is $(b-a)^2(1+(b+a)^2)$. This cannot be a perfect square, since $1+(b+a)^2$ cannot be: $x^2+1$ is a perfect square only when $x=0$.